A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

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Q.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

1 decreases by a factor of 2

2 remains the same

3 increases by a factor of 2

4 increases by a factor of 4

Ans.
(1)

When the capacitor is charged by a battery of potential $V$ , then energy stored in the capacitor,

$U_{i} = \frac{1}{2} C V^{2}$ ...(i)

When the battery is removed and another identical uncharged capacitor is connected in parallel

Common potential, $V' = \frac{C V}{C + C} = \frac{V}{2}$

Then the energy stored in the capacitor,

$U_{f} = \frac{1}{2} (2 C) {(\frac{V}{2})}^{2} = \frac{1}{4} C V^{2}$ ...(ii)

$∴$ $From eqns. (i) and (ii), U_{f} = \frac{U_{i}}{2}$

It means the total electrostatic energy of resulting system will decrease by a factor of 2.

Solution

Q.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

1 decreases by a factor of 2

2 remains the same

3 increases by a factor of 2

4 increases by a factor of 4

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Solution

Q. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

1 decreases by a factor of 2

2 remains the same

3 increases by a factor of 2

4 increases by a factor of 4

Want Us to Email you About Special Offers & Updates?

Q.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]